Algebra Examples

4

$4$ ,

8

$8$ ,

16

$16$ ,

32

$32$ ,

64

$64$

Step 1

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by

2

$2$ gives the next term. In other words,

a_{n} = a_{1} r^{n - 1}

$a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence:

r = 2

$r = 2$

Step 2

This is the form of a geometric sequence.

a_{n} = a_{1} r^{n - 1}

$a_{n} = a_{1} r^{n - 1}$

Step 3

Substitute in the values of

a_{1} = 4

$a_{1} = 4$ and

r = 2

$r = 2$ .

a_{n} = 4 \cdot 2^{n - 1}

$a_{n} = 4 \cdot 2^{n - 1}$

Step 4

Multiply

4 \cdot 2^{n - 1}

$4 \cdot 2^{n - 1}$ .

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Step 4.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

a_{n} = 2^{2} \cdot 2^{n - 1}

$a_{n} = 2^{2} \cdot 2^{n - 1}$

Step 4.2

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

a_{n} = 2^{2 + n - 1}

$a_{n} = 2^{2 + n - 1}$

Step 4.3

Subtract

1

$1$ from

2

$2$ .

a_{n} = 2^{n + 1}

$a_{n} = 2^{n + 1}$

a_{n} = 2^{n + 1}

$a_{n} = 2^{n + 1}$

Step 5

Substitute in the value of

n

$n$ to find the

n

$n$ th term.

a_{6} = 2^{(6) + 1}

$a_{6} = 2^{(6) + 1}$

Step 6

Add

6

$6$ and

1

$1$ .

a_{6} = 2^{7}

$a_{6} = 2^{7}$

Step 7

Raise

2

$2$ to the power of

7

$7$ .

a_{6} = 128

$a_{6} = 128$

4, 8, 16, 32, 64

$4, 8, 16, 32, 64$

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